467,858 Members | 1,657 Online
Bytes | Developer Community
New Post

Home Posts Topics Members FAQ

Post your question to a community of 467,858 developers. It's quick & easy.

Recursive zipping of Directories in Windows

Hi

I am a python novice and I am trying to write a python script (most of
the code is borrowed) to Zip a directory containing some other
directories and files. The script zips all the files fine but when it
tries to zip one of the directories it fails with the following
error:
"IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'"

The script I am using is:

import zipfile, os

def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for file in files:
file = os.path.join( directory, file )
# yes, the +1 is hacky...
archiveName = file[len(os.path.commonprefix( (root,
file) ))+1:]
zip.write( file, archiveName, zipfile.ZIP_DEFLATED )
print file
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa', 'c:\\aaa\\test.zip' )

I have tried to set the permissions on the folder, but when I check
the directory permissions it is set back to "Read Only"

Any suggestions?

Thanks
Johan Balt

Feb 1 '07 #1
  • viewed: 2563
Share:
5 Replies
Jandre wrote:
Hi

I am a python novice and I am trying to write a python script (most of
the code is borrowed) to Zip a directory containing some other
directories and files. The script zips all the files fine but when it
tries to zip one of the directories it fails with the following
error:
"IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'"

The script I am using is:

import zipfile, os

def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for file in files:
file = os.path.join( directory, file )
# yes, the +1 is hacky...
archiveName = file[len(os.path.commonprefix( (root,
file) ))+1:]
zip.write( file, archiveName, zipfile.ZIP_DEFLATED )
print file
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa', 'c:\\aaa\\test.zip' )

I have tried to set the permissions on the folder, but when I check
the directory permissions it is set back to "Read Only"

Any suggestions?

Thanks
Johan Balt
Couple of quick suggestions that may help:

1) don't use 'file' as a variable name. It will mask
the builtin file function. If it hasn't bitten you before
it will if you keep doing that.

2) If you put the target .zip file in the directory you are
backing what do you expect the program to do when it comes
to the file you are creating as you walk the directory? You
haven't done anything to 'skip' it.

3) Your commonprefix and +1 appears to result in same
information that the easier to use os.path.basename()
would give you. Double check me on that.

I don't see anything that references C:\\aaa\temp in your
code. Does it exist on your hard drive? If so does it
maybe contain temp files that are open? zipfile module
can't handle open files. You must use try/except to
catch these errors.

Hope info helps.

-Larry
Feb 1 '07 #2
On Feb 1, 9:39 pm, Larry Bates <larry.ba...@websafe.comwrote:
Jandre wrote:
Hi
I am a python novice and I am trying to write a python script (most of
the code is borrowed) to Zip a directory containing some other
directories and files. The script zips all the files fine but when it
tries to zip one of the directories it fails with the following
error:
"IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'"
The script I am using is:
import zipfile, os
def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for file in files:
file = os.path.join( directory, file )
# yes, the +1 is hacky...
archiveName = file[len(os.path.commonprefix( (root,
file) ))+1:]
zip.write( file, archiveName, zipfile.ZIP_DEFLATED )
print file
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa', 'c:\\aaa\\test.zip' )
I have tried to set the permissions on the folder, but when I check
the directory permissions it is set back to "Read Only"
Any suggestions?
Thanks
Johan Balt

Couple of quick suggestions that may help:

1) don't use 'file' as a variable name. It will mask
the builtin file function. If it hasn't bitten you before
it will if you keep doing that.

2) If you put the target .zip file in the directory you are
backing what do you expect the program to do when it comes
to the file you are creating as you walk the directory? You
haven't done anything to 'skip' it.

3) Your commonprefix and +1 appears to result in same
information that the easier to use os.path.basename()
would give you. Double check me on that.

I don't see anything that references C:\\aaa\temp in your
code. Does it exist on your hard drive? If so does it
maybe contain temp files that are open? zipfile module
can't handle open files. You must use try/except to
catch these errors.

Hope info helps.

-Larry
Thank you Larry.
I've changed the code as epr your advice. The code is now:

import zipfile, os

def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for f in files:
f = os.path.join( directory, f )
archiveName = os.path.basename(f)
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
print f
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )

I still get the same error:
Traceback (most recent call last):
File "C:\Python24\Lib\site-packages\pythonwin\pywin\framework
\scriptutils.py", line 310, in RunScript
exec codeObject in __main__.__dict__
File "C:\Python24\Scripts\dirZip.py", line 20, in ?
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )
File "C:\Python24\Scripts\dirZip.py", line 14, in toZip
os.path.walk( directory, walker, z )
File "C:\Python24\lib\ntpath.py", line 329, in walk
func(arg, top, names)
File "C:\Python24\Scripts\dirZip.py", line 12, in walker
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
File "C:\Python24\lib\zipfile.py", line 405, in write
fp = open(filename, "rb")
IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'

c:\\aaa\\temp is a directory in the directory I an trying to zip. I
want to use this script to back up my work once a day and would like
to
keep the directory structure as is. I can zip the files in c:\aaa\tem
fine so I guess that there aren't any open files in the directory.
Any more ideas?

Feb 4 '07 #3
On Feb 4, 12:42 pm, "Jandre" <jandre_b...@yahoo.co.ukwrote:
On Feb 1, 9:39 pm, Larry Bates <larry.ba...@websafe.comwrote:
Jandre wrote:
Hi
I am a python novice and I am trying to write a python script (most of
the code is borrowed) to Zip a directory containing some other
directories and files. The script zips all the files fine but when it
tries to zip one of the directories it fails with the following
error:
"IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'"
The script I am using is:
import zipfile, os
def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for file in files:
file = os.path.join( directory, file )
# yes, the +1 is hacky...
archiveName = file[len(os.path.commonprefix( (root,
file) ))+1:]
zip.write( file, archiveName, zipfile.ZIP_DEFLATED )
print file
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa', 'c:\\aaa\\test.zip' )
I have tried to set the permissions on the folder, but when I check
the directory permissions it is set back to "Read Only"
Any suggestions?
Thanks
Johan Balt
Couple of quick suggestions that may help:
1) don't use 'file' as a variable name. It will mask
the builtin file function. If it hasn't bitten you before
it will if you keep doing that.
2) If you put the target .zip file in the directory you are
backing what do you expect the program to do when it comes
to the file you are creating as you walk the directory? You
haven't done anything to 'skip' it.
3) Your commonprefix and +1 appears to result in same
information that the easier to use os.path.basename()
would give you. Double check me on that.
I don't see anything that references C:\\aaa\temp in your
code. Does it exist on your hard drive? If so does it
maybe contain temp files that are open? zipfile module
can't handle open files. You must use try/except to
catch these errors.
Hope info helps.
-Larry

Thank you Larry.
I've changed the code as epr your advice. The code is now:

import zipfile, os

def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for f in files:
f = os.path.join( directory, f )
archiveName = os.path.basename(f)
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
print f
os.path.walk( directory, walker, z )
z.close()
return zipFile

if __name__ == "__main__":
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )

I still get the same error:
Traceback (most recent call last):
File "C:\Python24\Lib\site-packages\pythonwin\pywin\framework
\scriptutils.py", line 310, in RunScript
exec codeObject in __main__.__dict__
File "C:\Python24\Scripts\dirZip.py", line 20, in ?
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )
File "C:\Python24\Scripts\dirZip.py", line 14, in toZip
os.path.walk( directory, walker, z )
File "C:\Python24\lib\ntpath.py", line 329, in walk
func(arg, top, names)
File "C:\Python24\Scripts\dirZip.py", line 12, in walker
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
File "C:\Python24\lib\zipfile.py", line 405, in write
fp = open(filename, "rb")
IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'

c:\\aaa\\temp is a directory in the directory I an trying to zip. I
want to use this script to back up my work once a day and would like
to
keep the directory structure as is. I can zip the files in c:\aaa\tem
fine so I guess that there aren't any open files in the directory.
Any more ideas?
Hi Jandre,

Your code is treating the directory as a file and trying to open it
and read its bytes to zip them. You'll need to differentiate between
files and directories.

You'll need to check out the Zip module to see how it expects files
that should be nested within folders. I believe you'll need to set the
archive name for the nested files to something like \\temp\\file.ext
etc.

-Josh

Feb 6 '07 #4
On Feb 6, 1:48 am, "joshbl...@gmail.com" <joshbl...@gmail.comwrote:
On Feb 4, 12:42 pm, "Jandre" <jandre_b...@yahoo.co.ukwrote:
On Feb 1, 9:39 pm, Larry Bates <larry.ba...@websafe.comwrote:
Jandre wrote:
Hi
I am a python novice and I am trying to write a python script (most of
the code is borrowed) to Zip a directory containing some other
directories and files. The script zips all the files fine but when it
tries to zip one of the directories it fails with the following
error:
"IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'"
The script I am using is:
import zipfile, os
def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for file in files:
file = os.path.join( directory, file )
# yes, the +1 is hacky...
archiveName = file[len(os.path.commonprefix( (root,
file) ))+1:]
zip.write( file, archiveName, zipfile.ZIP_DEFLATED )
print file
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa', 'c:\\aaa\\test.zip' )
I have tried to set the permissions on the folder, but when I check
the directory permissions it is set back to "Read Only"
Any suggestions?
Thanks
Johan Balt
Couple of quick suggestions that may help:
1) don't use 'file' as a variable name. It will mask
the builtin file function. If it hasn't bitten you before
it will if you keep doing that.
2) If you put the target .zip file in the directory you are
backing what do you expect the program to do when it comes
to the file you are creating as you walk the directory? You
haven't done anything to 'skip' it.
3) Your commonprefix and +1 appears to result in same
information that the easier to use os.path.basename()
would give you. Double check me on that.
I don't see anything that references C:\\aaa\temp in your
code. Does it exist on your hard drive? If so does it
maybe contain temp files that are open? zipfile module
can't handle open files. You must use try/except to
catch these errors.
Hope info helps.
-Larry
Thank you Larry.
I've changed the code as epr your advice. The code is now:
import zipfile, os
def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for f in files:
f = os.path.join( directory, f )
archiveName = os.path.basename(f)
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
print f
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )
I still get the same error:
Traceback (most recent call last):
File "C:\Python24\Lib\site-packages\pythonwin\pywin\framework
\scriptutils.py", line 310, in RunScript
exec codeObject in __main__.__dict__
File "C:\Python24\Scripts\dirZip.py", line 20, in ?
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )
File "C:\Python24\Scripts\dirZip.py", line 14, in toZip
os.path.walk( directory, walker, z )
File "C:\Python24\lib\ntpath.py", line 329, in walk
func(arg, top, names)
File "C:\Python24\Scripts\dirZip.py", line 12, in walker
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
File "C:\Python24\lib\zipfile.py", line 405, in write
fp = open(filename, "rb")
IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'
c:\\aaa\\temp is a directory in the directory I an trying to zip. I
want to use this script to back up my work once a day and would like
to
keep the directory structure as is. I can zip the files in c:\aaa\tem
fine so I guess that there aren't any open files in the directory.
Any more ideas?

Hi Jandre,

Your code is treating the directory as a file and trying to open it
and read its bytes to zip them. You'll need to differentiate between
files and directories.

You'll need to check out the Zip module to see how it expects files
that should be nested within folders. I believe you'll need to set the
archive name for the nested files to something like \\temp\\file.ext
etc.
In my experience, zip files don't contain nested folders, instead they
contain a 'flat' list of files. Files in subfolders are represented by
relative path names containing slashes and empty subfolders are
represented by relative paths ending in a slash,.for example
"File1.txt", "Folder1/File2.dat", "Folder2/Folder3/". This seems to
work for me! :-)
Feb 6 '07 #5
Jim
On Feb 6, 2:47 pm, "MRAB" <goo...@mrabarnett.plus.comwrote:
On Feb 6, 1:48 am, "joshbl...@gmail.com" <joshbl...@gmail.comwrote:
On Feb 4, 12:42 pm, "Jandre" <jandre_b...@yahoo.co.ukwrote:
On Feb 1, 9:39 pm, Larry Bates <larry.ba...@websafe.comwrote:
Jandre wrote:
Hi
I am a python novice and I am trying to write a python script (most of
the code is borrowed) to Zip a directory containing some other
directories and files. The script zips all the files fine but when it
tries to zip one of the directories it fails with the following
error:
"IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'"
The script I am using is:
import zipfile, os
def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for file in files:
file = os.path.join( directory, file )
# yes, the +1 is hacky...
archiveName = file[len(os.path.commonprefix( (root,
file) ))+1:]
zip.write( file, archiveName, zipfile.ZIP_DEFLATED )
print file
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa', 'c:\\aaa\\test.zip' )
I have tried to set the permissions on the folder, but when I check
the directory permissions it is set back to "Read Only"
Any suggestions?
Thanks
Johan Balt
Couple of quick suggestions that may help:
1) don't use 'file' as a variable name. It will mask
the builtin file function. If it hasn't bitten you before
it will if you keep doing that.
2) If you put the target .zip file in the directory you are
backing what do you expect the program to do when it comes
to the file you are creating as you walk the directory? You
haven't done anything to 'skip' it.
3) Your commonprefix and +1 appears to result in same
information that the easier to use os.path.basename()
would give you. Double check me on that.
I don't see anything that references C:\\aaa\temp in your
code. Does it exist on your hard drive? If so does it
maybe contain temp files that are open? zipfile module
can't handle open files. You must use try/except to
catch these errors.
Hope info helps.
-Larry
Thank you Larry.
I've changed the code as epr your advice. The code is now:
import zipfile, os
def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for f in files:
f = os.path.join( directory, f )
archiveName = os.path.basename(f)
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
print f
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )
I still get the same error:
Traceback (most recent call last):
File "C:\Python24\Lib\site-packages\pythonwin\pywin\framework
\scriptutils.py", line 310, in RunScript
exec codeObject in __main__.__dict__
File "C:\Python24\Scripts\dirZip.py", line 20, in ?
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )
File "C:\Python24\Scripts\dirZip.py", line 14, in toZip
os.path.walk( directory, walker, z )
File "C:\Python24\lib\ntpath.py", line 329, in walk
func(arg, top, names)
File "C:\Python24\Scripts\dirZip.py", line 12, in walker
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
File "C:\Python24\lib\zipfile.py", line 405, in write
fp = open(filename, "rb")
IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'
c:\\aaa\\temp is a directory in the directory I an trying to zip. I
want to use this script to back up my work once a day and would like
to
keep the directory structure as is. I can zip the files in c:\aaa\tem
fine so I guess that there aren't any open files in the directory.
Any more ideas?
Hi Jandre,
Your code is treating the directory as a file and trying to open it
and read its bytes to zip them. You'll need to differentiate between
files and directories.
You'll need to check out the Zip module to see how it expects files
that should be nested within folders. I believe you'll need to set the
archive name for the nested files to something like \\temp\\file.ext
etc.

In my experience, zip files don't contain nested folders, instead they
contain a 'flat' list of files. Files in subfolders are represented by
relative path names containing slashes and empty subfolders are
represented by relative paths ending in a slash,.for example
"File1.txt", "Folder1/File2.dat", "Folder2/Folder3/". This seems to
work for me! :-)- Hide quoted text -

- Show quoted text -
What's the matter with PKZIP? It's been around forever.

Feb 6 '07 #6

This discussion thread is closed

Replies have been disabled for this discussion.

Similar topics

1 post views Thread by gusmeister | last post: by
4 posts views Thread by Elmo Watson | last post: by
reply views Thread by Benjamin Bittner | last post: by
6 posts views Thread by reidarT | last post: by
3 posts views Thread by Gabe Matteson | last post: by
By using this site, you agree to our Privacy Policy and Terms of Use.